- #1

JulienB

- 408

- 12

## Homework Statement

Hi everybody! I'm struggling to solve the following problem:

Let ##< \cdot, \cdot >## be an inner product on the vector space ##X##, and ##|| \cdot ||## is the norm generated by the inner product. Prove that the function ##x \in X \mapsto ||x||^2 \in \mathbb{R}## is differentiable.

## Homework Equations

I copy here what I have in my script:

##\exists L \in \mathbb{L} (X;Y): \lim\limits_{y \to 0} \frac{f(x + y) - f(x) - L y}{||y||} = 0 \implies f## is differentiable at ##x## and ##L## is unique. When ##\exists f'(x) \forall x \in M \implies f## is differentiable and ##f': M \to \mathbb{L} (X;Y)## is the derivative (##M \subseteq X## is an open set, ##\mathbb{L} (X;Y)## is the set of all multilinear continuous functions mapping from ##X## to ##Y##).

## The Attempt at a Solution

Well... This is very abstract to me. I assume that the norm generated by the inner product is:

##||x|| := \sqrt{<x,x>}##

which leads me to

##||x||^2 = <x,x>##.

Inserting this in the definition for differentiability gives me:

##\lim\limits_{y \to 0} \frac{f(x + y) - f(x) - L y}{||y||} = \lim\limits_{y \to 0} \frac{<x+y,x+y> - <x,x> - L y}{\sqrt{<y,y>}}##

##= \lim\limits_{y \to 0} \frac{<x,x+y> + <y,x+y> - <x,x> - L y}{\sqrt{<y,y>}}##

##= \lim\limits_{y \to 0} \frac{<x,x> + <x,y> + <y,x> + <y,y> - <x,x> - L y}{\sqrt{<y,y>}}##

##= \lim\limits_{y \to 0} \frac{2<x,y> + <y,y> - L y}{\sqrt{<y,y>}}##

To begin, I'm not sure that's the way to go. I didnt get any other bright idea though. Notice that in the last line, I used the property of the scalar product ##<x,y> = <y,x>## which should hold if we are in ##\mathbb{R}^n##. It is not clear to me if that is the case, since no information is given about ##X##. What would you guys say?

Anyway I am kind of stuck then. I kind somewhat rewrite the expression I got as:

##\lim\limits_{y \to 0} \frac{2<x,y> + <y,y> - L y}{\sqrt{<y,y>}}##

##= \lim\limits_{y \to 0} \sqrt{<y,y>} + \frac{2<x,y> - L y}{\sqrt{<y,y>}}##

##= \lim\limits_{y \to 0} \frac{2<x,y> - L y}{\sqrt{<y,y>}}##

and then I am quite stuck again. Can I somehow differentiate using L'Hopital? That seems pretty unlikely.Thank you guys in advance for your answers, I appreciate it.Julien.